#### Find the minimum value of the function f(x)=3^[(x^2-2)^3 +8]

Solution:    We have ,

$f(x)=3^{(x^{2}-2)^{3}+8}$

The  function   $f(x)=3^{\phi (x)}$  takes on the minimum value at the

same point  as the function $\phi (x)$.

Hence ,

$\phi (x)=x^{6}-6x^{4}+12x^{2}=x^{2}[(x^{2}-3)^{2}+3].$

Whence it is clear that the funtion $\phi (x)$ attains the minimum value

at $x=0$ . that is why the minimum value of the function  $f(x)$ is

equal to $3^{0}=1$